# Reading for Class 7: Interference and Diffraction of Light Waves

## Interfering Waves

Two traveling waves can interfere with each other.  You may remember how to add sine waves from a trig class long ago.  You merely add the functions at each individual point.  For example, consider the three cases illustrated below.

In case (a), the two waves match up exactly.  We say they are "in-phase", since everywhere Wave 1 has a crest Wave 2 also has a crest, and everywhere Wave 1 has a trough, Wave 2 also has a trough, etc.  When these two waves add, the result is a sine wave with twice the amplitude as one of the original waves.  The waves in this case exhibit complete constructive interference.

In case (b), the waves are "180º out of phase", which means that they are off by one-half cycle.  Every crest in Wave 1 matches a trough in Wave 2, and vice-versa.  Adding these two waves yields an amplitude of zero.  This case illustrates complete destructive interference.

Case (c) illustrates a situation somewhere between complete constructive and complete destructive interference.  The two waves do not line up completely, but they do not cancel either.  The result is a sine wave with an amplitude lower than twice the amplitude of one wave.  The exact value of the amplitude will depend on how out-of-phase the two waves are.  If the waves are nearly aligned, the amplitude of the sum will be larger than if the two waves are nearly canceling.

The difference between the distances traveled by the two waves is called the path difference, and it is represented by d on the diagram.  The waves in case (a) needn't have traveled the exact same distance to line up; if one wave has traveled one wavelength more than the other has, the two waves would still line up.  They would similarly align if one has traveled two or three or any integer number of wavelengths further than the other has.  If the path difference is equal to an integer number of wavelengths, the two waves will experience complete constructive interference, and their amplitudes will add.

Similarly, the path difference for the waves in case (b) needn't be just l/2; if one has traveled three half-wavelengths more than the other has, they would still cancel.  They would also cancel if one has traveled any other half-integer number of wavelengths further than the other has.  If the path difference is equal to a half- integer number of wavelengths, the two waves will experience complete destructive interference, their amplitudes will cancel, and the sum will always have zero amplitude.  Thus our conditions on the path difference d  for interference are as follows:

d = ml  (m = 0, ±1, ±2, . . .)   for constructive interference.
d = (m + 1/2)l   (m= 0, ±1, ±2, . . .)  for destructive interference.

## Diffraction of Waves through an Aperture

When a wave hits an opening, or aperture, it spreads out on the other side of the aperture.  This phenomenon is known as diffraction.  We can represent the wave as either a series of crests (red solid lines) and troughs (red dashed lines), or as a ray pointing the direction of wave motion (blue arrow).  As the wave passes through the aperture, the crests and troughs take a circular shape, and the direction of wave motion becomes radial.  One consequence of diffraction is that waves can travel around corners.  Observation of light's ability to travel around corners and diffract helped bolster the wave theory of light.

## Interference through a Double Slit

When a wave passes through two slits, it diffracts at both openings.  The circular waves emitted by one opening can then interfere with the circular waves coming out the other opening.  If the wave from the bottom opening (yellow) has traveled an integer number of wavelengths further (or less) than the wave from the top opening (green), the result will be complete constructive interference (pink dots).  Similarly, complete destructive interference (brown dots) will occur at locations where the wave from the bottom opening has traveled a half-integer number of wavelengths further (or less) than the wave from the top opening has.

If the waves are light waves, spots of constructive interference will be bright, and spots of destructive interference will be dark.  A screen represented by the black line in the above diagram would therefore display regions of high-intensity light dimming into regions of zero intensity before brightening into high-intensity regions again.  This collection of bright and dark regions is called an "interference pattern" for light.

## Interference through a Single Slit

Light traveling through a single slit can also produce an interference pattern.  One can think of the sources of the interfering rays as being different parts of the slit.  To visualize this case, it is helpful to think of the light as "rays" (green and yellow).  A ray incident on the lower part of the slit will diverge, as will a ray incident on the upper part of the slit.  The resulting diverging waves will interfere, in a manner similar to the interference of a double slit.  Bright spots, called maxima, will appear when the path difference between the interfering rays is an integer number of wavelengths.  Dark minima will appear when the path difference is a half-integer number of wavelengths.

The pattern produced by a single slit is similar to, but not identical to, the pattern produced by a double slit.  To distinguish between the two situations, we call the double-slit pattern an interference pattern, and the single-slit pattern a diffraction pattern.  This distinction is somewhat misleading, since both effects depend upon both diffraction (the spreading of light) and interference (combining of light), but it is commonly used so you should be aware of it.

## Interference through a Circular Aperture

If a circular aperture is used instead of a narrow slit, the diffraction pattern is circular, looking something like the figure below.  We will use such a pattern to determine the wavelengths of two lasers in class.  To do so, we first define two distances: D is the distance between the aperture and the screen, and y is the distance between the center of the pattern and the first minimum.  The right-hand side of the figure below shows the top view of the setup.  From this drawing, we can see that the angle q is related to y and D by
tan q = y/D.

Deriving the next equation is beyond the scope of this course.  Suffice it to say that the path difference d between the rays coming from different parts of the aperture and arriving at the point of the first minimum is

d = (d sin q)/2.44.

Here d is the diameter of the aperture, and the factor of 2.44 comes from mathematical functions describing circular geometry.  As mentioned above, when the light rays cancel to produce the first minimum, this path difference must be l/2.  Setting these two expressions for path difference equal to each other, we get

l/2 = (d sin q)/2.44
l = 2(d sin q)/2.44 = (d sin q)/1.22.

For small angles, sin q is approximately equal to tan q, so

sin q ~ tan q = y/D = 1.22 l/d.

Thus the location y of the first minimum depends on the aperture size d, the distance D between the aperture and the screen, and the wavelength l.

## Application

In an optical storage device such as a CD-ROM or DVD, laser light is focused by a lens onto the surface of the disk.  The lens, while focusing the light, also serves as a circular aperture.  Thus, the light coming out will produce an interference pattern on the disk.  If more than one bit falls inside the central bright spot, the bits cannot be resolved and information will be lost.  This limit of resolvability is described by the Rayleigh criteria:

sin q £ 1.22 l/d.

The bit must be no smaller than a circle of radius r, where

r/D = tan q~ 1.22 l/d.
r ~ 1.22 lD/d

Theoretically, one could decrease this minimum bit size in three different ways:  using a laser with a shorter wavelength l, decreasing the distance D between the lens and the disk, or using a lens with a larger diameter d.  Moving the lens closer to the disk or using a larger lens might seem the simplest ways to reduce the diffraction pattern, but D and d are in practice tightly constrained.  Laser light passing through the lens will be focused to the smallest spot at the focal point of the lens, so the disk-to-lens distance D should equal the focal length of the lens.  Thus D is determined by the lens and cannot be arbitrarily varied.  Manufacturing constraints on lenses also limit the width d of the lens.  Photographers are familiar with the f-number, or f-stop of lenses.  This represents the ratio of focal length D over diameter d of the lens.  Lenses with large f-numbers such as f/8 or f/16 are relatively easy to produce.  As the f-number decreases, the difficulty increases, and lenses with f-numbers less than 1 are very rare.  Thus the ratio D/d will be greater than 1(and probably more like 5) for the mass-produced lenses in CD drives.  This leaves us with the option of reducing the laser wavelength as the most promising way to decrease the minimum bit size in optical storage.

One of the reason's DVDs can hold more data than CD-ROMs is that they use a shorter wavelength of light, so the resulting diffraction pattern on the disk is smaller.  Bits may then each occupy less space, so more bits fit on a disk.  Your handout from Ron White's book discusses additional reasons that DVDs store more data than CD-ROMs.

## Summing it Up

This reading assignment has a lot of information stuffed into a few pages of text.  Many of the equations have been included to prove the few relationships we will need.  You will not be asked to repeat any of those derivations, but you should take away with you the following key concepts:

 - Light passing through a circular aperture is diffracted and produces a ringed diffraction pattern. - For two objects to be resolved with such light, they must not both lie within the central maximum. - The size y of the central maximum can be reduced in three ways:  decreasing the distance D between the slit and the screen, increasing the diameter d of the aperture, and decreasing the wavelength l. - In practical applications (like CD-ROMs), the aperture size d and distance between aperture and screen (data) D are severly constrained.  Thus adjusting the wavelength becomes the most effective way to decrease the size of the diffraction pattern and thereby resolve smaller objects.